Final Answer
$\frac{\sqrt{2z^2+7z+3}}{\sqrt{z^2-9}}$
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Step-by-step Solution
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1
The limit of a constant is just the constant
$\sqrt{\frac{2z^2+7z+3}{z^2-9}}$
2
The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
$\frac{\sqrt{2z^2+7z+3}}{\sqrt{z^2-9}}$
Final Answer
$\frac{\sqrt{2z^2+7z+3}}{\sqrt{z^2-9}}$